# Prinicipal Ideal Domain (P.I.D.) | Discrete Mathematics

**Prerequisite :** Rings in Discrete Mathematics

**Introduction :**

Algebraic Structure : A non-empty set G equipped with 1 or more binary operations is called algebraic structure.**Example – **

- (N,+) where N is a set of natural numbers and
- (R, *) R is a set of real numbers.

Here ‘ * ‘ specifies a multiplication operation.

**RING : **

An algebraic structure that sets processing of two binary operations simultaneously is needed to form a Ring. A non-empty set R together with the operations multiplication & addition (Usually) is called a ring if :

1. (R,+) is an Abelian Group (satisfies G1, G2, G3, G4 & G5) 2. (R, *) is a Semi Group. (satisfies G1 & G2) 3. Multiplication is distributive over addition : (a) Left Distributive : a*(b+c) = (a*b) + (a*c) ; ∀ a, b, c ∈ R (b) Right Distributive : (b+c)*a = (b*a) + (c*a) ; ∀ a, b, c ∈ R

**GROUP –**

An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called a Group if the binary operation “o” satisfies the following properties :

G1. Closure : a ∈ G ,b ∈ G => aob ∈ G ; ∀ a,b ∈ G

G2. Associativity : (aob)oc = ao(boc) ; ∀ a,b,c ∈ G.

G3. Identity Element : There exists e in G such that aoe = eoa = a ; ∀ a ∈ G (Example – For addition, identity is 0)

G4. Existence of Inverse : For each element a ∈ G ; there exists an inverse(a-1)∈ G such that : aoa-1 = a-1oa = e.**Abelian Group –**

An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called an abelian Group if it is a group (i.e. , it satisfies G1, G2, G3 & G4) and additionally satisfies :

G5 : Commutative: aob = boa ∀ a,b ∈ G**Semi Group –**

An algebraic structure (G , o) where G is a non-empty set & ‘o’ is a binary operation defined on G is called a semi-group if it satisfies only 2 properties : G1 (closure ) & G2 (Associativity).

We usually write the ring structure as : (R, +, *) or simply R.

Note : Additive identity 0 is unique & is called the zero element of the Ring R.**Commutative Ring –**

(R,+, *) is commutative : it means that the multiplication (*) is commutative.

**Integral Domain : **

A ring (R, +, *) is called an integral domain of :

1. (R,+,*) is commutative. 2. (R,+,*) is a ring with unit element. 3. It is a ring without zero divisors.

**(R,+, *) is commutative –**

means that the multiplication (*) is commutative.**(R,+,*) is a ring with a unit element –**

It means that there exists a unit element, say 1∈ R, such that :

a*1 = 1*a = a ∀ a ∈ R**R is a ring without zero divisors –**

a*b = 0 =>a = 0 OR b = 0 where a, b ∈ R

**Principal Ideal :**

Let (R,+, *) is a commutative ring with identity 1.

Let a ∈R, then the set = { ra : r ∈ R is an ideal } called the Principal Ideal generated by a.

**Principal Ideal Domain (P.I.D.) :**

A ring (R,+, *) is called a principal ideal domain if :

- R is an integral domain.
- Every ideal in R is principal.

If every one-sided ideal of a ring is ideal, it is termed a primary ideal ring. The principal ideal domain is a principal ring with no zero divisors.**Note :** integrally closed domains ⊂ integral domains ⊂ commutative rings ⊂ rings

**Q. Showing that every ****field**** is a P.I.D.****Solution.** Let F is a field. Therefore, F is an integral domain too. Also, F would have some unity element : a*1 = 1*a = a ∀ a ∈ F. So, F is an integral domain with unity.

Every field has only 2 ideals. So F has 2 ideals : {0} & F where –

(i) {0} = 0*F

(ii) F = 1*F

So, F has only 2 ideals & they can be expressed in the form : { f*a : f ∈ R is an ideal and a ∈ F }

So, every F is a P.I.D.

Note : The converse may not be true.

**Q. Show that Z,the ring of integers, is a P.I.D.**

Answer. We know that Z, the set of integers, is an integral domain.

Let J be an ideal in Z. We show J is a principal ideal.**Case 1. –** If J = {0}, then it is the principal ideal & hence the result.**Case 2. –** If J ≠ {0}, Let 0 ≠ x ∈ J, then -x = (-1) x ∈ J for some positive x.

Hence, J contains at least one positive integer. Let a be the smallest positive integer in J.

We claim, J = { ra : r ∈Z }

For x ∈ J , using the division algorithm,

x = qa + r ; 0 ≤ r ≤ a ; q ∈ Z

But J is an ideal and a ∈ J, q ∈ Z

Hence, qa ∈ J and x – qa ∈ J ⇒ r ∈ J.

But, a is the smallest positive integer in J satisfying 0 ≤ r ≤ a. Hence, we must have r = 0.

So, x = qa , i.e.,

J = { qa : q ∈ Z}

Hence Z is a P.I.D.

**Q. Prove : A PID is a unique factorization domain.****Proof :** The reverse inclusion relation in the set of nonzero ideals is well-founded in classical logic.

Let A denote the subset of ideals (a) that are products of a finite number(possibly zero) of maximal principal ideals . If every (t) correctly containing (x) can be factored into maximals for each valid ideal (x)(0), then so can (x). (Either (x) is maximal/irreducible, or it factors as (s)(t), where both s and t are non-units; by hypothesis, (s) and (t) factor into maximals, so does (x)). As a result, because A is an inductive set, it contains every ideal (x)(0), i.e., x may be factored into irreducible.

For the factorization’s uniqueness, we first note that if p is irreducible and p|ab, then p|a or p|b. (Because R/(p) is a field and hence a fortiori an integral domain, if ab≡0modp is true, then a≡0modp or b≡0modp is true as well.). p1 divides one of the irreducible if q_{i} ,p_{1}p_{2}…p_{m}=q_{1}q_{2}…q_{n} are two factorizations into irreducible of the same element, then , in which case (p_{1})=(q_{i}) and each is a unit times the other, meaning we can cancel p_{1} on both sides and argue by induction.

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